Twenty years ago, my teacher of ODE said that the solution of $y'(x)=y(x)$ is ehm, trivailly, $y(x)=0$ or $y(x)=ke^x$. Since then, I have tried everything (MVT, tweaking with $e$ number...), but I couldn't prove that a solution that holds $y(0)=0$ is necessarily $y\equiv 0$. A good link would succify. Thanks.
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2Let $z(t) = e^{-t} y(t)$ then note that $z'(t) = 0$ and so $z(t) = z(0)$. The latter follows from the MVT. – copper.hat Feb 15 '23 at 02:51
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Seriously you defferentate and integrate for a single point?? – ajotatxe Feb 15 '23 at 03:06
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I don't understand your comment. What single point and what integration? – copper.hat Feb 15 '23 at 03:07
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Why is $z'(t)=0$? – ajotatxe Feb 15 '23 at 03:08
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1Are you familiar with differentiating a product? – copper.hat Feb 15 '23 at 03:09
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I am a Mathematician – ajotatxe Feb 15 '23 at 03:10
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1Then differentiate $z$ as defined above. – copper.hat Feb 15 '23 at 03:10
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2Since $y’=y$ that quantity is $0$. – Vivaan Daga Feb 15 '23 at 03:16
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See https://math.stackexchange.com/a/660453/72031 – Paramanand Singh Feb 15 '23 at 03:37
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The general solution to $y'=y$ is $y=ke^x$ where $k$ is any constant (see this question). There's no need to treat the case $k=0$ separately and say “$y=0$ or $y=ke^x$”. – Hans Lundmark Feb 15 '23 at 11:22
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If $y$ is differentiable and satisfies $y'=y$ then let $z(t) = e^{-t} y(t)$. Differentiating gives $z'(t) = e^{-t} (y'(t)-y(t)) = 0$. Hence $z'(t) = 0$ and so, by the mean value theorem, $z(t) = z(0)$.
Since $y(t) =e^t z(t)$, we see that $y(t) = e^t y(0)$.
copper.hat
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The other answer says why 0 is a solution. To see why it's the only solution: the Picard–Lindelöf theorem roughly says that if $f(x, y)$ is continuous in $x$ and Lipschitz in $y$, then there's a unique solution for some closed interval around your initial condition. In the case of your ODE the relevant function is $f(x, y) = y$, which fulfills these requirements, so the solution is unique.
Davis Yoshida
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I don´t think so. TVM and Bolzano, even Liouville, doesn't work. The problem is that the set of solutions if usually given as a fact. They ussualy take $y=y'$ and then write $y'/y=1$, with no reagrds about if $y$ can have zeroes. – ajotatxe Feb 15 '23 at 03:02
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@Shinrin-Yoku TBH I'm not a big diff eq person so I just pulled up the first existence and uniqueness theorem I could remember. – Davis Yoshida Feb 15 '23 at 03:30
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1@Shinrin-Yoku: It might be overkill in this particular case, but it's still a useful general fact worth knowing that if the hypotheses for Picard–Lindelöf are satisfied, and if $f(y^)=0$, then the equilibrium solution $y(x)=y^$ is the unique solution to the initial value problem $y'=f(y)$, $y(0)=y^*$. – Hans Lundmark Feb 15 '23 at 11:26
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@ajotatxe: If you solve it using $e^{-x}$ as an integrating factor instead of using the method of separation of variables, there's no need to worry about $y$ being zero. – Hans Lundmark Feb 15 '23 at 11:28
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HINT
You can solve it and study the obtained solution.
In the present case, the solution to the proposed ODE is given by $y(x) = ke^{x}$.
Now it remains the question: when we assume that $y(0) = 0$, what does it imply?
Átila Correia
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